holt algebra 1 8-3 factoring x 2 + bx + c 1.factor quadratic trinomials of the form x 2 + bx + c....
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Holt Algebra 1
8-3 Factoring x2 + bx + c
1.Factor quadratic trinomials of the form x2 + bx + c.
Objective
2. Factor quadratic trinomials of the form x2 + bx + c.
3. Factor four terms by grouping(front 2, back 2)
Holt Algebra 1
8-3 Factoring x2 + bx + c
Example 1A: Factoring by Using the GCFFactor each polynomial. Check your answer.
–14x – 12x2
Check –2x(7 + 6x)
Multiply to check your answer.
The product is the original polynomial.
–14x – 12x2
Holt Algebra 1
8-3 Factoring x2 + bx + c
Example 1B
Factor each polynomial. Check your answer.
8x4 + 4x3 – 2x2
8x4 = 2 2 2 x x x x4x3 = 2 2 x x x2x2 = 2 x x
2 x x = 2x2
4x2(2x2) + 2x(2x2) –1(2x2)
2x2(4x2 + 2x – 1)
Check 2x2(4x2 + 2x – 1)
8x4 + 4x3 – 2x2
The GCF of 8x4, 4x3 and –2x2 is 2x2.
Multiply to check your answer.
The product is the original polynomial.
Write terms as products using the GCF as a factor.
Use the Distributive Property to factor out the GCF.
Find the GCF.
Holt Algebra 1
8-3 Factoring x2 + bx + c
Example 2
Factor each expression.
a. 4s(s + 6) – 5(s + 6)
4s(s + 6) – 5(s + 6) The terms have a common binomial factor of (s + 6).
(4s – 5)(s + 6) Factor out (s + 6).
b. 7x(2x + 3) + (2x + 3)
7x(2x + 3) + (2x + 3)
7x(2x + 3) + 1(2x + 3)
(2x + 3)(7x + 1)
The terms have a common binomial factor of (2x + 3).
(2x + 1) = 1(2x + 1)
Factor out (2x + 3).
Holt Algebra 1
8-3 Factoring x2 + bx + c
You may be able to factor a polynomial by grouping. When a polynomial has four terms, you can make two groups and factor out the GCF from each group.
Holt Algebra 1
8-3 Factoring x2 + bx + c
Example 3A: Factoring by Grouping
Factor each polynomial by grouping. Check your answer.
6h4 – 4h3 + 12h – 8
(6h4 – 4h3) + (12h – 8)
2h3(3h – 2) + 4(3h – 2)
2h3(3h – 2) + 4(3h – 2)
(3h – 2)(2h3 + 4)
Group terms that have a common number or variable as a factor.
Factor out the GCF of each group.
(3h – 2) is another common factor.
Factor out (3h – 2).
Holt Algebra 1
8-3 Factoring x2 + bx + c
Example 3A Continued
Factor each polynomial by grouping. Check your answer.
Check (3h – 2)(2h3 + 4) Multiply to check your solution.
3h(2h3) + 3h(4) – 2(2h3) – 2(4)
6h4 + 12h – 4h3 – 8
The product is the original polynomial.
6h4 – 4h3 + 12h – 8
Holt Algebra 1
8-3 Factoring x2 + bx + c
Example 43B: Factoring by Grouping
Factor each polynomial by grouping. Check your answer.
5y4 – 15y3 + y2 – 3y
(5y4 – 15y3) + (y2 – 3y)
5y3(y – 3) + y(y – 3)
5y3(y – 3) + y(y – 3)
(y – 3)(5y3 + y)
Group terms.
Factor out the GCF of each group.
(y – 3) is a common factor.
Factor out (y – 3).
Holt Algebra 1
8-3 Factoring x2 + bx + c
Example 4: Factoring with Opposites
Factor 2x3 – 12x2 + 18 – 3x
2x3 – 12x2 + 18 – 3x
(2x3 – 12x2) + (18 – 3x)
2x2(x – 6) + 3(6 – x)
2x2(x – 6) + 3(–1)(x – 6)
2x2(x – 6) – 3(x – 6)
(x – 6)(2x2 – 3)
Group terms.
Factor out the GCF of each group.
Simplify. (x – 6) is a common factor.
Factor out (x – 6).
Write (6 – x) as –1(x – 6).
Holt Algebra 1
8-3 Factoring x2 + bx + c
Lesson Quiz: Part 1
Factor each polynomial. Check your answer.
1. 16x + 20x3
2. 4m4 – 12m2 + 8m
Factor each expression (by grouping).
3. 3y(2y + 3) – 5(2y + 3)
(2y + 3)(3y – 5)
4m(m3 – 3m + 2)
4x(4 + 5x2)
4. 2x3 + x2 – 6x – 3
5. 7p4 – 2p3 + 63p – 18
(2x + 1)(x2 – 3)
(7p – 2)(p3 + 9)
Holt Algebra 1
8-3 Factoring x2 + bx + c
(x + 2)(x + 5) = x2 + 7x + 10
Notice that when you multiply (x + 2)(x + 5), the constant term in the trinomial is the product of the constants in the binomials.
When you multiply two binomials, multiply:
First terms
Outer terms
Inner terms
Last terms
Remember!
Holt Algebra 1
8-3 Factoring x2 + bx + c
The guess and check method is usually not the most efficient method of factoring a trinomial. Look at the product of (x + 3) and (x + 4).
(x + 3)(x +4) = x2 + 7x + 12
x2 12
3x4x
The coefficient of the middle term is the sum of 3 and 4. The third term is the product of 3 and 4.
Holt Algebra 1
8-3 Factoring x2 + bx + c
Holt Algebra 1
8-3 Factoring x2 + bx + c
Example 1: Factoring x2 + bx + c
x2 + 6x + 5
Factor each trinomial. Check your answer.
(x + )(x + ) b = 6 and c = 5; look for factors of 5 whose sum is 6.
Factors of 5 Sum
1 and 5 6 The factors needed are 1 and 5.
(x + 1)(x + 5)
Check (x + 1)(x + 5) = x2 + x + 5x + 5 Use the FOIL method.The product is the
original polynomial.= x2 + 6x + 5
Holt Algebra 1
8-3 Factoring x2 + bx + c
Example 2: Factoring x2 + bx + c
Factor each trinomial. Check your answer.
x2 – 8x + 15
b = –8 and c = 15; look for factors of 15 whose sum is –8.
The factors needed are –3 and –5 .
Factors of –15 Sum–1 and –15 –16
–3 and –5 –8
(x – 3)(x – 5)
Check (x – 3)(x – 5 ) = x2 – 3x – 5x + 15 Use the FOIL method.The product is the
original polynomial.= x2 – 8x + 15
(x + )(x + )
Holt Algebra 1
8-3 Factoring x2 + bx + c
Example 3: Factoring x2 + bx + c
Factor each trinomial.
x2 + x – 20
(x + )(x + ) b = 1 and c = –20; look for factors of –20 whose sum is 1. The factor with the greater absolute value is positive.
The factors needed are +5 and –4.
Factors of –20 Sum
–1 and 20 19 –2 and 10 8
–4 and 5 1
(x – 4)(x + 5)
Holt Algebra 1
8-3 Factoring x2 + bx + c
Factor each trinomial. Check your answer.
Example 4
x2 + 2x – 15
(x + )(x + )
Factors of –15 Sum
–1 and 15 14 –3 and 5 2
(x – 3)(x + 5)
b = 2 and c = –15; look for factors of –15 whose sum is 2. The factor with the greater absolute value is positive.
The factors needed are –3 and 5.
Holt Algebra 1
8-3 Factoring x2 + bx + c
Lesson Quiz: Part I
Factor each trinomial.
1. x2 – 11x + 30
2. x2 + 10x + 9
3. x2 – 6x – 27
4. x2 + 14x – 32
(x + 16)(x – 2)
(x – 9)(x + 3)
(x + 1)(x + 9)
(x – 5)(x – 6)